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Sorry for sounding stupid, but I am really new to jQuery so I see the code you provided and that is much appreciated for sure, thank you for getting back to me. I been trying to fix this one thing for 2 days and just can't get it to work.

So, now my question is what part of my script should fire after that check occurs? I played around with it a little bit, but it didn't exactly. It left all panels open and nothing would fire when clicked. So...I'm confused here. ugh.

Note, it doesn't matter where the script is placed in this example, but if you do decide to leave it at the bottom then you will get better loading performance by removing the $(document).ready() piece in the code.
See best practices for speeding up your web site

wow, thank you very much for helping out here. This is great. I put it in my file and it well basically almost works completely.

The only thing that doesn't seem to work is when i click a tab and its open, if i click it again, it should close. Right now it doesn't do anything. I will check with my people to see if that works for them or not. I see it in the code the line that tells the dd:visible to slide up but that doesn't fire for some reason.

One last hurdle to cross here:
Now when i mouseover a button, it lights up, when i click a button, it shows the on state, what state would the button be in when you click the same button a second time?

So you click the button once to open it up and show the ACTIVE image, so how can u make it show another image when you close it?

i'm using a different script, but i don't know what state it would be in for the 2nd click?